@t.b.: Surely not...the question here is "does there exist $a, b, c, d\in\mathbb{Q}$ not all zero such that $a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}=0$", while the question you link to boils down to $\sqrt{2}\sqrt{3}=\sqrt{6}$, which is an illegal move here...
–
user1729Jan 6 '12 at 15:13

2

@t.b.: Not quite (thought not unrelated). For example, $\sqrt{6}$ is certainly in $\mathbb{Q}(\sqrt{2},\sqrt{3})$, but isn't in the $\mathbb{Q}$-vector space generated by $1$, $\sqrt{2}$, and $\sqrt{3}$.
–
Cam McLemanJan 6 '12 at 15:15

Actually, @user1729, Bill Dubuque's answer in the linked question does in fact answer this question (and some more). And the fact that $b\sqrt{2} + c\sqrt{3} \not\in Q(\sqrt{6})$ (as demonstrated below by Martin) is basically how one shows (in an elementary way) that $Q(\sqrt{2},\sqrt{3}) \neq Q(\sqrt{6})$ (Alvaro's proof in the linked question). That is to say I agree with t.b. that this question is an abstract duplicate of the other, though whether it should be closed as such is another matter.
–
Willie WongJan 6 '12 at 15:39

2

@WillieWong: Yeah, I had a similar paragraph written before I decided to scrap it in favor of "though not unrelated." In any case, I more or less agree, though it's unclear to me the extent to which the existence of repeatable answers implies that the questions themselves are duplicate.
–
Cam McLemanJan 6 '12 at 16:56

3 Answers
3

I have no doubt that many slick solutions will be given here. I'll try to post an elementary one; which uses perhaps the most straightforward approach I can imagine. (The only things that are needed are some algebraic manipulation and basic properties of rational numbers; such as that the only solution of $x^2=6y^2$ in $\mathbb Q$ are $x=y=0$.)

This is equivalent to showing that if
$$a+b\sqrt6=c\sqrt3+d\sqrt2$$
then $a=b=c=d=0$.

By squaring both sides of the above equation we get
$$
\begin{align*}
a^2+6b^2+2ab\sqrt6&=3c^2+2d^2+2cd\sqrt6\\
2(ab-cd)\sqrt6=3c^2+2d^2-a^2-6b^2
\end{align*}
$$
Since $a,b,c,d\in\mathbb Q$, this implies
$$
\begin{align*}
ab-cd&=0\\
3c^2+2d^2-a^2-6b^2&=0
\end{align*}
$$
which is the same as
$$
\begin{align*}
ab&=cd\\
3c^2+2d^2&=a^2+6b^2
\end{align*}
$$

Suppose that $b\ne0$, $c\ne0$. (I'll leave the solution of these cases to the reader.) Then we can rewrite the first equation as $\frac ac = \frac db = x$, where $x\in\mathbb Q$. Now the second equation becomes
$$
\begin{align*}
3c^2+2x^2b^2&=x^2c^2+6b^2\\
x^2(2b^2-c^2)&=3(2b^2-c^2)\\
(x^2-3)(2b^2-c^2)&=0
\end{align*}$$
This implies that $x^2=3$ or $2b^2=c^2$. None of them has non-zero solutions in rational numbers.

@Martin Sleziak:I think I have got your idea.Thanks.And I have read a version of proof that it first gives any two of them and any three of them are linearly independent.After having read your solution,I want to ask whether the version I read before is useful or not.Is it necessary to do like that?And one more question,is there any proof that based on any other different theory and can be presented in an elementary way like yours?
–
AndylangJan 6 '12 at 16:21

1

@Andylang: The parts which I "left for the reader" are using the fact that three/two of them are linearly independent. (But you probably don't need to do all possible triples.) For different proofs just check the comments under your question.
–
Martin SleziakJan 6 '12 at 16:41

REMARK $\ $ By induction, the lemma easily generalizes to algebraic extensions generated by adjoining $\rm\:n\:$ square-roots, see my post here on Besicovic's Theorem, which includes references to generalizations by Mordell and Siegel. These results are elementary special cases of the Galois theory of Kummer extensions.

Observe that $a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6} = 0$ if and only if $(a + b\sqrt{2}) + (c + d\sqrt{2})\sqrt{3} = 0$. Hence it is enough to show that $\sqrt{3}$ and $1$ are independent over $\mathbb{Q}(\sqrt{2})$, but this is equivalent to showing that $\sqrt{3}\notin\mathbb{Q}(\sqrt{2})$, or that $\sqrt{3}$ cannot be written as $a + b\sqrt{2}$. This is obvious (square both sides and play with the result).